1:
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2-1-9n^2}{\sqrt{n^2-1}-3n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{-8n^2-1}{\sqrt{n^2-1}-3n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2\left(-8-\dfrac{1}{n^2}\right)}{n\left(\sqrt{1-\dfrac{1}{n^2}}-3\right)}=\lim\limits_{n\rightarrow\infty}-\dfrac{8}{1-3}\cdot n=\lim\limits_{n\rightarrow\infty}4n=+\infty\)
2:
\(\lim\limits_{n\rightarrow\infty}\sqrt{4n^2+5}+n\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{4n^2+5-n^2}{\sqrt{4n^2+5}-n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{3n^2+5}{\sqrt{4n^2+5}-n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2\left(3+\dfrac{5}{n^2}\right)}{n\left(\sqrt{4+\dfrac{5}{n^2}}-1\right)}\)
\(=\lim\limits_{n\rightarrow\infty}n\cdot\left(\dfrac{3}{\sqrt{4}-1}\right)=+\infty\)